Fundamental Actuarial Mathematics/Mortality Models

Learning objectives

The Candidate will understand key concepts concerning parametric and non-parametric mortality models for individual lives.

Learning outcomes

The Candidate will be able to:

1. Understand parametric survival models, life tables, and the relationships between them.
2. Given a parametric survival model, calculate survival and mortality probabilities, the force of mortality function, and curtate and complete moments of the future lifetime random variable.
3. Identify and apply standard actuarial notation for future lifetime distributions and moments, including select and ultimate functions.
4. Given a life table, calculate survival and mortality probabilities, the force of mortality function, and curtate and complete moments of the future lifetime random variable, using appropriate fractional age assumptions where necessary.
5. Understand and apply select life tables.
6. Identify common features of population mortality curves.

Survival distributions

In the model discussed in this chapter, it describes the length of survival (or time until death) of an individual. Thus, the time-until-death random variable will be the basic building block.

Age-at-death random variable

In this section, we will discuss a special case for the time-until-death random variable, in which the time until death is applicable to newborn (i.e. people aged zero). We denote this kind of random variable by $T_{0}$. We can observe that $T_{0}$ also represents the age at death, since the age is counted starting from the beginning of life.

Since time-until-death random variable is describing time, it is a continuous random variable. Also, time is nonnegative, so the support (or "domain") of time-until-death random variable is $[0,\infty )$.

To describe the time until death for newborn, we need to determine the distribution of $T_{0}$ completely. There are several ways to do this.

• cumulative distribution function (cdf): $F_{0}(t)=\mathbb {P} (T_{0}\leq t)$
• probability density function (pdf): $f_{0}(t)=F'_{0}(t)$ if $F_{0}(t)$ is differentiable.
• survival function
• force of mortality

You should have learnt about cdf and pdf when learning about probability, but may not have learnt about survival function and force of mortality. Thus, we will discuss them here.

Survival function

As suggested by the name "survival function", we may guess that this function is somewhat related to survival. This is actually true. Take the time-until-death random variable $T_{0}$ as an example, when the newborn survives for, say, $t$ units of time, what is its probability? It is $\mathbb {P} (T_{0}>t)$ (or $\mathbb {P} (T_{0}\geq t)$, but since $T_{0}$ is continuous, it does not matter). This probability corresponding to the input $t$ is actually the survival function, which is defined below:

Definition. (Survival function) The survival function of a random variable $X$ is $S_{X}(x)=\mathbb {P} (X>x)$.

Remark.

• As a corollary, $S_{X}(x)=1-F_{X}(x)$.

• We can see that $F_{0}(t)$ gives the probability for the newborn to die within $t$ units of time.
• Since $F_{X}$ is a nondecreasing function, $S_{X}$ is a nonincreasing function.

• For the survival function of $T_{0}$, we denote it by $S_{0}(t)$, which equals $1-F_{0}(t)$.

• If $t<0$, $S_{0}(t)=\mathbb {P} (X>t)=1$ since the support is a subset of $\{X>t\}$, so this event "covers" the entire support already.

Example. Recall that the cdf of the exponential distribution with rate $\lambda$ is $F(x)=1-e^{-\lambda x}$. Thus, the survival function of the exponential distribution with rate $\lambda$ is $S(x)=e^{-\lambda x}$.

Force of mortality

In financial mathematics, you should have learnt about force of interest, which can be interpreted as the relative rate of change of the amount function, and it is given by ${\frac {A'(t)}{A(t)}}$ in which the notation has its usual meaning in financial mathematics. Why do we call it force of interest? This is because the interest refers to an increase (or positive change) in the amount.

We may guess that the force of mortality is defined similarly, in the sense that it can also be interpreted as the relative rate of change of something. We know that the interest refers to change in amount, but what change does mortality refer to? Since mortality means the state of being susceptible to death, it refers to the decrease (or negative change) in survival rate, and the "higher" the mortality, the larger decrease in survival rate. Recall that the survival function is related to survival rate (probability of surviving for a certain time) in some sense. So, we can make use of survival function to define mortality.

However, there is a difference between force of interest and force of mortality, namely for force of interest, interest refers to an increase in amount, while for force of mortality, mortality refers to a decrease in survival rate. So the changes are in opposite direction, and thus if we define force of mortality in an exact analogous manner, its value will be negative (the relative rate of change will be negative). To make the force of mortality positive, we can define the force of mortality as follows:

Definition. (Force of mortality) The force of mortality of a random variable $X$ is $\mu _{X}(x)={\frac {-S_{X}'(x)}{S_{X}(x)}}$.

Remark.

• For force of mortality of the time-until-death random variable $T_{0}$, it is denoted by $\mu _{t}$, which equals ${\frac {-S'_{0}(t)}{S_{0}(t)}}$.

• If $t<0$, since $S_{0}(t)=1$ always, $S'_{0}(t)=0$ and thus $\mu _{t}=0$
• If $t\geq 0$, since $S_{0}(t)$ is nonincreasing, its derivative is nonpositive, and so the negative of it is nonnegative. Also, $S_{0}(t)\geq 0$ (because the survival function is a probability). Therefore, $\mu _{t}\geq 0$.

• The force of mortality of $T_{0}$ at time $t$ shows the relative rate of decrease in the survival rate of the newborn at time $t$.

Example. Show that $\mu _{t}={\frac {f_{0}(t)}{S_{0}(t)}}$, assuming $F_{0}(t)$ is differentiable.

Solution: $${\begin{aligned}\mu _{t}&={\frac {-S_{0}'(t)}{S_{0}(t)}}\\&={\frac {-(1-F_{0}(t))'}{S_{0}(t)}}\\&={\frac {-(-f_{0}(t))}{S_{0}(t)}}&{\text{since }}F'_{0}(t)=f'_{0}(t)\\&={\frac {f_{0}(t)}{S_{0}(t)}}.\\\end{aligned}}$$

Remark.

• We can interpret $f_{0}(t)/S_{0}(t)$ (roughly) as the conditional probability of the newborn to die "instantaneously" at time $t$, given that the newborn survives for $t$ time units, since $f_{0}(t)/S_{0}(t)\approx \mathbb {P} (t<T_{0}\leq t+\Delta t)/\mathbb {P} (T_{0}>t)=\mathbb {P} (\underbrace {t<T_{0}\leq t+\Delta t} _{{\text{die within the next }}\Delta t{\text{ time units}}}|T_{0}>t)$ ($\Delta t$ is small and close to zero).
• This interpretation is intuitive, since the "instantaneous" death rate at time $t$ may be interpreted as the relative rate of decrease of the survival rate at time $t$ (when the "instantaneous" death rate is high at that time point, the relative survival rate at that time point diminishes a lot).

Exercise.

What is the force of mortality $\mu _{X}$ of the exponential distribution with rate $\lambda$?

	$\lambda$
	$-\lambda$
	1
	-1

After that, we will introduce some propositions related to the survival function and force of mortality.

Proposition. $S_{0}(t)=\exp \left(-\int _{0}^{t}\mu _{x}\,dx\right).$

Proof. $${\begin{aligned}\exp \left(-\int _{0}^{t}\mu _{x}\,dx\right)&=\exp \left({\color {darkgreen}-}\int _{0}^{t}{\frac {{\color {darkgreen}-}S'_{0}(x)}{S_{0}(x)}}\,dx\right)\\&=\exp \left(\int _{0}^{t}{\frac {\color {red}S'_{0}(x)}{S_{0}(x)}}\,{\color {red}dx}\right)\\&=\exp \left(\int _{0}^{t}{\frac {1}{S_{0}(x)}}\,{\color {red}d(S_{0}(x))}\right)\\&=\exp \left([\ln |\underbrace {S_{0}(x)} _{\geq 0}|]_{0}^{t}\right)\\&=\exp \left(\ln(S_{0}(t))-\ln(\underbrace {S_{0}(0)} _{=1})\right)\\&=\exp \left(\ln(S_{0}(t))-\underbrace {\ln 1} _{=0}\right)\\&=\exp \left(\ln(S_{0}(t))\right)\\&=S_{0}(t),\end{aligned}}$$ as desired.

$\Box$

Proposition. $\int _{0}^{\infty }\mu _{t}\,dt=\infty .$

Proof. For simplicity of presentation, we will have some abuse of notations (the infinity is in limit sense) in the following, but the reasoning is still understandable. $${\begin{aligned}\int _{0}^{\infty }\mu _{t}\,dt&=[\ln(S_{0}(t))]_{0}^{\infty }\\&=\ln S_{0}(\infty )-\ln S_{0}(0)\\&=\ln 0\\&=\infty .\end{aligned}}$$

$\Box$

Example. Given that $T_{0}$ follows the uniform distribution with support $[0,100]$, what is the force of mortality of $T_{0}$, $\mu _{t}$?

Solution: We have $S_{0}(t)=1-F_{0}(t)=1-{\frac {t-0}{100-0}}=1-{\frac {t}{100}}$. So, $$\mu _{t}={\frac {-S_{0}'(t)}{S_{0}(t)}}={\frac {-(1-t/100)'}{1-t/100}}={\frac {1/100}{(100-t)/100}}={\frac {1}{100}}\cdot {\frac {100}{100-t}}={\frac {1}{100-t}}.$$

Exercise.

Suppose instead the force of mortality of $T_{0}$, $\mu _{t}=t/100$. What is the probability for the newborn to die within 10 years?

	0.368
	0.393
	0.607
	0.632

Future lifetime of a life aged x

Now, we extend our discussion from future lifetime of a life aged zero (a newborn) to a life aged $x$ ($x\geq 0$). For simplicity of presentation, we denote a life aged $x$ by $(x)$.

Remark.

• When we say "a life aged $x$", we mean the life is aged exactly $x$, i.e. the life just reaches age $x$ (birthday of the life), but not, say aged $x+0.5$, $x+0.9$, etc, which are usually also referred as "aged $x$" in our daily life.

Similarly, we denote the future lifetime of $(x)$ by $T_{x}$ (recall that we denote the future lifetime of $(0)$ (newborn) by $T_{0}$). We define the distribution of $T_{x}$ mathematically (and quite naturally) as the conditional distribution of $T_{0}-x$, given that $T_{0}>x$.

To understand this, consider the following reasoning: Refer to the following timeline:

                  death
    x       T_x   |
|---------|-------v
-------------------------
0         x                 t
|-----------------|
      T_0

We can observe that $T_{x}=T_{0}-x$ if $T_{0}>x$ (or $T_{0}\geq x$, but since $T_{0}$ is continuous, it does not matter). So, if $T_{0}>x$, then $T_{x}=T_{0}-x$.

On the other hand, if $T_{0}<x$, we have the following timeline:

      death       
    x   |       
|-------v-|
-------------------------
0         x                 t
|-------|
   T_0

In this case, $T_{x}$ does not exist, since the person does not survive for $x$ years, and thus will never be age $x$, so there is not $(x)$, and therefore there is not $T_{x}$, future lifetime of $(x)$. This shows the necessity of the condition $T_{0}>x$.

From this definition, we have $\mathbb {P} (T_{x}\leq t)=\mathbb {P} (T_{0}-x\leq t|T_{0}>x)$, $\mathbb {P} (T_{x}>t)=\mathbb {P} (T_{0}-x>t|T_{0}>x)$, etc.. This is quite important since it is the basis for the calculations of probabilities related to $T_{x}$.

For the pdf, cdf and survival function of $T_{x}$, we have similar notations as follows:

• $f_{x}(t)$: pdf of $T_{x}$
• $F_{x}(t)$: cdf of $T_{x}$
• $S_{x}(t)$: survival function of $T_{x}$

In particular, we have some special actuarial notations for the cdf and survival function, as follows:

• $_{t}q_{x}=F_{x}(t)=\mathbb {P} (T_{x}\leq t)$
• $_{t}p_{x}=S_{x}(t)=\mathbb {P} (T_{x}>t)=1-{}_{t}q_{x}$

In actuarial notations, "$q$" often refers to something related to death, while "$p$" often refers to something related to survival. In this context, this holds since $_{t}q_{x}$ refers to the probability for $(x)$ to die within $t$ time units, and $_{t}p_{x}$ refers to the probability for $(x)$ to survive for $t$ time units.

For simplicity, if $t=1$, we write $_{1}q_{x}$ as $q_{x}$ and $_{1}p_{x}$ as $p_{x}$.

Using the relationship between $T_{x}$ and $T_{0}$, we can develop some useful formulas for $_{t}p_{x}$ and $_{t}q_{x}$, as follows:

Proposition. $_{t}p_{x}={\frac {S_{0}(t+x)}{S_{0}(x)}}$ and $_{t}q_{x}=1-{\frac {S_{0}(t+x)}{S_{0}(x)}}$.

Proof. First, we have $_{t}p_{x}=\mathbb {P} (T_{x}>t)=\mathbb {P} (T_{0}-x>t|T_{0}>x)={\frac {\mathbb {P} (T_{0}>t+x\cap T_{0}>x)}{\mathbb {P} (T_{0}>x)}}={\frac {\mathbb {P} (T_{0}>t+x)}{\mathbb {P} (T_{0}>x)}}={\frac {S_{0}(t+x)}{S_{0}(x)}}$, in which $\{T_{0}>t+x\}\cap \{T_{0}>x\}=\{T_{0}>t+x\}$ since $t+x>x$ ($t\geq 0$), and so $T_{0}>t+x\implies T_{0}>x$, and thus $\{T_{0}>t+x\}$ is a subset of $\{T_{0}>x\}$.

It follows that $_{t}q_{x}=1-{}_{t}p_{x}=1-{\frac {S_{0}(t+x)}{S_{0}(x)}}$.

$\Box$

We can also express the pdf of $T_{x}$ as follows:

Proposition. $f_{x}(t)={}_{x}p_{t}\mu _{x+t}$.

Proof. We have $f_{x}(t)={\frac {d}{dt}}{}_{t}q_{x}=\left(1-{\frac {S_{0}(t+x)}{S_{0}(x)}}\right)'={\frac {-S'_{0}(t+x)}{S_{0}(x)}}={\frac {\color {darkgreen}S_{0}(t+x)}{S_{0}(x)}}\cdot {\frac {-S'_{0}(t+x)}{\color {darkgreen}S_{0}(t+x)}}={}_{t}p_{x}\mu _{x+t}.$

$\Box$

Remark.

• Intuitively (and roughly), $_{t}p_{x}$ gives the probability for $(x)$ to survive for $t$ time units and after that, $(x)$ becomes $(x+t)$, and $\mu _{x+t}$ gives the probability for $(x+t)$ to die "instantaneously" at time $x+t$, given that the person survives to $x+t$ time units. Multiplying $_{t}p_{x}$ and $\mu _{x+t}$, it means the same as the (rough) interpretation of $f_{x}(t)$: $(x)$ die very shortly after (or "exactly at") time $x+t$.

Example. It is given that the survival function of newborn is $S_{0}(t)=1-{\frac {t}{10}},\quad 0\leq t\leq 10$.

(a) Calculate $_{2}p_{1}$ and $q_{2}$. Hence, determine whether $_{2}p_{1}=p_{2}$.

(b) Calculate $\mu _{3}$. Hence, calculate $f_{2}(1)$.

Solution:

(a) $_{2}p_{1}={\frac {S_{0}(3)}{S_{0}(1)}}={\frac {0.7}{0.9}}\approx 0.778$ and $q_{2}=1-{\frac {S_{0}(3)}{S_{0}(2)}}=1-{\frac {0.7}{0.8}}=0.125$. Since $p_{2}=1-q_{2}=0.875$, $_{2}p_{1}\neq p_{2}$.

(b) Since $S'_{0}(t)=-{\frac {1}{10}}$, $\mu _{3}={\frac {-S'_{0}(3)}{S_{0}(3)}}={\frac {-(-1/10)}{0.7}}\approx {\frac {1}{7}}$. So, $f_{2}(1)={}_{1}p_{2}\mu _{3}=(0.875)(1/7)=0.125$.

Exercise.

Find $f_{x}(t)$.

	$f_{x}(t)={\frac {1}{10-x}}$
	$f_{x}(t)={\frac {1}{10-t}}$
	$f_{x}(t)={\frac {1}{10-2t}}$
	$f_{x}(t)={\frac {t}{10-x}}$

We have a special notation for the probability for $(x)$ to die between ages $x+t$ and $x+t+u$ ($x,t,u\geq 0$), namely $_{t|u}q_{x}$ (we use "$q$" here since this is related to death). Thus, we have by definition $_{t|u}q_{x}=\mathbb {P} (t<T_{x}<t+u))$. We have the following proposition for another formula of $_{t|u}q_{x}$.

Proposition. $_{t|u}q_{x}={}_{t}p_{x}({}_{u}q_{x+t})$.

Proof. $${\begin{aligned}_{t|u}q_{x}&=\mathbb {P} (t<T_{x}<t+u)\\&=\mathbb {P} (T_{x}\leq t+u)-\mathbb {P} (T_{x}\leq t)\\&={}_{t+u}q_{x}-_{t}q_{x}\\&=(1-_{t+u}p_{x})-(1-_{t}p_{x})\\&={}_{t}p_{x}-_{t+u}p_{x}\\&={}_{t}p_{x}-{\frac {S_{0}(x+t+u)}{S_{0}(x)}}\\&={}_{t}p_{x}-{\frac {S_{0}(x+t+u)}{\color {darkgreen}S_{0}(x+t)}}\left({\frac {\color {darkgreen}S_{0}(x+t)}{S_{0}(x)}}\right)\\&={}_{t}p_{x}-{}_{u}p_{x+t}({}_{t}p_{x})\\&={}_{t}p_{x}(1-{}_{u}p_{x+t})\\&={}_{t}p_{x}(_{u}q_{x+t})\\\end{aligned}}$$

$\Box$

Remark.

• For proving formulas like this, it is generally better to change all "$q$" to "$p$" in the intermediate steps since "$p$" is usually better to work with than "$q$".
• To understand this more intuitively, $_{t}p_{x}$ can be interpreted as the probability for $(0)$ to survive for $x+t$ time units, given that $(0)$ survives for $x$ time units, and $_{u}q_{x+t}$ can be interpreted as the probability for $(0)$ to die within $x+t+u$ time units, given that $(0)$ survives for $x+t$ time units. Therefore, multiplying these two probability yields the probability for $(0)$ to die within $x+t+u$ time units, and survive for $x+t$ time units, given that $(0)$ survives for $x$ time units.

• This argument corresponds to the ${\frac {S_{0}(x+t+u)}{S_{0}(x)}}={\frac {S_{0}(x+t+u)}{\color {darkgreen}S_{0}(x+t)}}{\frac {\color {darkgreen}S_{0}(x+t)}{S_{0}(x)}}$ in the above proof.
• If we denote the above blue event as ${\color {blue}A}$, orange event as ${\color {darkorange}B}$, and purple event as ${\color {purple}C}$, we can represent the above argument using probability notations: $\mathbb {P} ({\color {blue}A}|{\color {darkorange}B})\mathbb {P} ({\color {purple}C}|{\color {blue}A})=\mathbb {P} ({\color {blue}A}\cap {\color {purple}C}|{\color {darkorange}B})$.

• When you try to prove this equality, you can observe that this is equivalent to the $\underbrace {\frac {S_{0}(x+t+u)}{S_{0}(x)}} _{\mathbb {P} ({\color {blue}A}\cap {\color {purple}C}|{\color {darkorange}B})}=\underbrace {\frac {S_{0}(x+t+u)}{\color {darkgreen}S_{0}(x+t)}} _{\mathbb {P} ({\color {purple}C}|{\color {blue}A})}\underbrace {\frac {\color {darkgreen}S_{0}(x+t)}{S_{0}(x)}} _{\mathbb {P} ({\color {blue}A}|{\color {darkorange}B})}$ in the above proof.

• Similarly, we denote $_{t|1}q_{x}$ by $_{t|}q_{x}$ for simplicity.

Example. It is given that the survival function of newborn is $S_{0}(t)=e^{-t},\quad t\geq 0$.

(a) Calculate $_{3|}q_{2}$.

(b) Calculate $_{2|}q_{3}$.

(c) Are the answers in (a) and (b) the same?

Solution:

(a) $_{3|}q_{2}={}_{3}p_{2}(q_{5})={\frac {e^{-5}}{e^{-2}}}\left(1-{\frac {e^{-6}}{e^{-5}}}\right)={\frac {e^{-5}}{e^{-2}}}-{\frac {e^{-6}}{e^{-2}}}$

(b) $_{2|}q_{3}={}_{2}p_{3}(q_{5})={\frac {e^{-5}}{e^{-3}}}\left(1-{\frac {e^{-6}}{e^{-5}}}\right)={\frac {e^{-5}}{e^{-3}}}-{\frac {e^{-6}}{e^{-3}}}$

(c) They are not the same.

Exercise.

1 Which of the following expression(s) represent(s) the probability for $(5)$ to die between age 10 and 12?

	$_{7}p_{5}-{}_{5}p_{5}$
	$_{5}p_{5}-{}_{7}p_{5}$
	$_{5}q_{5}-{}_{7}q_{5}$
	$_{7}q_{5}-{}_{5}q_{5}$
	$_{10|12}q_{5}$

2 Which of the following expression(s) represent(s) the probability for $(5)$ to survive between age 10 and 12?

	$_{7}p_{5}$
	$_{5}p_{5}-{}_{7}q_{5}$
	$_{7}p_{5}-{}_{5}q_{5}$
	$_{7}p_{5}+{}_{5}q_{5}$
	$1-{}_{10|12}q_{5}$

Curtate-future-lifetime of a life aged x

The curtate-future-lifetime is just like the future lifetime in previous sections, except that it is discrete.

Definition. (Curtate-future-lifetime) The curtate-future-lifetime of $(x)$, denoted by $K_{x}$, is $\lfloor T_{x}\rfloor$, which is the floor function of $T_{x}$.

Remark.

• Hence, the support of $K_{x}$ is set of all nonnegative integers.

Similarly, we would like to completely determine the distribution of $K_{x}$, as in the case for $T_{x}$. We can do this using cdf or probability mass function (pmf). Its pmf is given by the following proposition.

Proposition. (Pmf of $K_{x}$) The pmf of $K_{x}$ is $_{k|}q_{x}={}_{k}p_{x}(q_{x+k})$.

Proof. The pmf of $K_{x}$ is $${\begin{aligned}\mathbb {P} (K_{x}=k)&=\mathbb {P} (k\leq T_{x}<k+1)\\&=\mathbb {P} (k<T_{x}<k+1)&{\text{since }}T_{x}{\text{ is continuous}}\\&={}_{k|}q_{x}&{\text{ by definition}}\\&={}_{k}p_{x}(q_{x+k})&{\text{by proposition}}.\\\end{aligned}}$$

$\Box$

Proposition. (Cdf of $K_{x}$) The cdf of $K_{x}$ is $\mathbb {P} (K_{x}\leq k)={}_{k+1}q_{x}$.

Proof. The cdf of $K_{x}$ is $${\begin{aligned}\mathbb {P} (K_{x}\leq k)&=\sum _{j=0}^{k}{}_{j|}q_{x}\\&={}_{0|}q_{x}+{}_{1|}q_{x}+\dotsb +{}_{k|}q_{x}\\&=\mathbb {P} ((x){\text{ dies between ages }}0{\text{ and }}1)+\mathbb {P} ((x){\text{ dies between ages }}1{\text{ and }}2)+\dotsb +\mathbb {P} ((x){\text{ dies between ages }}k{\text{ and }}k+1)\\&=\mathbb {P} ((x){\text{ dies between ages }}0{\text{ and }}k+1)\\&=\mathbb {P} ((x){\text{ dies within }}k+1{\text{ years}})\\&={}_{k+1}q_{x}.\end{aligned}}$$

$\Box$

Example. It is given that the survival function of newborn is $S_{0}(t)={\frac {100-t}{100}},\quad 0\leq t\leq 100$.

(a) Calculate the probability for $(20)$ to die within 10 years by considering $T_{20}$.

(b) Calculate the probability for $(20)$ to die within 10 years by considering $K_{20}$.

(c) Which probability, that in (a) or that in (b), is larger?

Solution:

(a) The probability is $\mathbb {P} (T_{20}\leq 10)={}_{10}q_{20}=1-{\frac {S_{0}(30)}{S_{0}(20)}}=1-{\frac {0.7}{0.8}}=0.125$.

(b) The probability is $\mathbb {P} (K_{20}\leq 10)={}_{11}q_{20}=1-{\frac {S_{0}(31)}{S_{0}(20)}}=1-{\frac {0.69}{0.8}}=0.1375$

(c) The probability in (b) is larger.

Exercise.

1 Which of the following is (are) the survival function of $K_{x}$?

	$_{k+1}p_{x}$
	$_{k+1}q_{x}$
	$_{k}p_{x}$
	$_{k}q_{x}$

2 Calculate the probability for $(20)$ to survive for 20 years by considering $K_{20}$.

	0.65
	0.7375
	0.75
	0.9875

Life tables

In a life table, the values of $q_{x}$ and other functions for different (integer) ages $x$ are tabulated. The values are assumed to be based on the survival distribution discussed in previous sections. In this section, we will discuss more functions appearing in a life table.

In previous sections, we have discussed time-until-death random variable for one person, and we will consider multiple people here. Suppose there are $\ell _{0}$ newborns. Let the indicator function $$\mathbf {1} _{j}(x)={\begin{cases}1,&{\text{if life }}j{\text{ survives to age }}x;\\0,&{\text{otherwise}}.\end{cases}}$$ Also let ${\mathcal {L}}(x)$ be the sum of all such indicator functions $\mathbf {1} _{j}(x)$, i.e. ${\mathcal {L}}(x)=\sum _{j=1}^{\ell _{0}}\mathbf {1} _{j}(x)$. We can interpret ${\mathcal {L}}(x)$ as the number of survivors to age $x$ for the $\ell _{0}$ newborns.

We denote the expected value of ${\mathcal {L}}(x)$ by $\ell _{x}$.

Proposition. $\ell _{x}=\ell _{0}S_{0}(x)$.

Proof. Since $\mathbb {E} [\mathbf {1} _{j}(x)]=1\cdot \mathbb {P} ({\text{life }}j{\text{ survives to age }}x)+0\cdot \mathbb {P} ({\text{life }}j{\text{ does not survive to age }}x)=\mathbb {P} ({\text{life }}j{\text{ survives to age }}x)=S_{0}(x)$ (this is true for each life $j$, since the survival distribution for the future lifetime of different lifes are assumed to be the same), $\ell _{x}$ equals $\mathbb {E} [{\mathcal {L}}(x)]=\mathbb {E} \left[\sum _{j=1}^{\ell _{0}}\mathbf {1} _{j}(x)\right]=\sum _{j=1}^{\ell _{0}}\mathbb {E} [\mathbf {1} _{j}(x)]=\sum _{j=1}^{\ell _{0}}\underbrace {S_{0}(x)} _{{\text{constant with respect to }}j}=\ell _{0}S_{0}(x)$.

$\Box$

As a corollary, $\ell '_{x}=(\ell _{0}S_{0}(x))'=\ell _{0}S'_{0}(x)$ ($\ell _{0}$ is constant with respect to $x$). Also, ${\frac {-\ell '_{x}}{\ell _{x}}}={\frac {-\ell _{0}S'_{0}(x)}{\ell _{0}S_{0}(x)}}={\frac {-S'_{0}(x)}{S_{0}(x)}}=\mu _{x}$. Also, we can use $\ell _{x}$ to calculate probabilities like $_{t}p_{x}$ and $_{t}q_{x}$, as follows: $_{t}p_{x}={\frac {S_{0}(x+t)}{S_{0}(x)}}={\frac {\ell _{0}S_{0}(x+t)}{\ell _{0}S_{0}(x)}}={\frac {\ell _{x+t}}{\ell _{x}}}$, and thus $_{t}q_{x}=1-{}_{t}p_{x}=1-{\frac {\ell _{x+t}}{\ell _{x}}}$. In a later section in which selection age is involved in the life table (select table), we will use these formulas to calculate these probabilities from such life table, to incorporate the effect of selection.

We have discussed about the number of survivors to age $x$, and we will discuss the "opposite thing" in the following, namely the number of deaths to age $x$ (i.e. between age 0 and $x$), or in general, between age $x$ and $x+n$.

We denote the expected value of such number of deaths by $_{n}d_{x}$.

Proposition. $_{n}d_{x}=\ell _{x}-\ell _{x+n}$.

Proof. We can define another indicator function for this context similarly (with value 1 if life $j$ dies between age $x$ and age $x+n$ and 0 otherwise). Then, the expected value of each indicator function equals the probability for one of the newborns to die between $x$ and $x+n$, which is $S_{0}(x)-S_{0}(x+n)$. With similar reasoning as in above, we have $_{n}d_{x}=\ell _{0}(S_{0}(x)-S_{0}(x+n))=\ell _{0}S_{0}(x)-\ell _{0}S_{0}(x+n)=\ell _{x}-\ell _{x+n}$.

$\Box$

Remark. Similarly, we write $_{1}d_{x}$ as $d_{x}$ for simplicity.

Apart from the life table functions $\ell _{x}$ and $_{n}d_{x}$ which are related to the expectation of number of survivors and deaths respectively, we will also discuss two more life table functions, that is related to the expectation of lifetime.

Example. Suppose $\ell _{0}=1000$ and the survival function for newborn is $S_{0}(x)=1-{\frac {x}{80}},\quad 0\leq x\leq 80$.

(a) What is $\ell _{80}$?

(b) What are $_{50}d_{30}$ and $_{50}d_{20}$? Are they equal?

Solution:

(a) $\ell _{80}=\ell _{0}S_{0}(80)=1000(0)=0$. (That is, the expected number of survivors to age 80 is 0.)

(b) $_{50}d_{30}=\ell _{30}-\ell _{80}=1000(1-30/80)-0=625$ and $_{50}d_{20}=\ell _{20}-\ell _{70}=1000(1-20/80)-1000(1-70/80)=625$. They are equal.

Remark.

• For (b), indeed, $_{50}d_{x}=625$ for each $x\in [0,30]$, since $_{50}d_{x}=1000(1-x/80)-1000(1-(50+x)/80)=1000(50/80)=625$ for each $x\in [0,30]$.

Exercise. It is given that $\ell _{x}=100-x,\quad 0\leq x\leq 100$.

1 What is $_{10}p_{30}$?

	0.143
	0.333
	0.667
	0.857

2 Calculate (i) $_{25}d_{30}$; (ii) $_{25}d_{80}$.

	(i) 25; (ii) 20
	(i) 20; (ii) 20
	(i) 20; (ii) 25
	(i) 25; (ii) 25

3. (a) Calculate ${}_{n}q_{x}$ in this case ($0\leq n\leq 100-x$). (b) Hence, show that $\ell _{x}{}_{n}q_{x}={}_{n}d_{x}$ in all cases. (c) Using (b) directly, show that $\ell _{x}{}_{n}p_{x}=\ell _{x+n}$ in all cases.

Solution

(a) $${}_{n}q_{x}=1-{\frac {S_{0}(x+n)}{S_{0}(x)}}=1-{\frac {\ell _{0}S_{0}(x+n)}{\ell _{0}S_{0}(x)}}=1-{\frac {\ell _{x+n}}{\ell _{x}}}=1-{\frac {100-x-n}{100-x}}.$$

(b)

Proof. From (a), we can observe that ${}_{n}q_{x}=1-{\frac {\ell _{x+n}}{\ell _{x}}}$ in all cases. Hence, $\ell _{x}{}_{n}q_{x}=\ell _{x}\left(1-{\frac {\ell _{x+n}}{\ell _{x}}}\right)=\ell _{x}-\ell _{x+n}={}_{n}d_{x}$ where the last equality follows from a previous proposition.

$\Box$

(c)

Proof. From (b), we have $$\ell _{x}{}_{n}q_{x}={}_{n}d_{x}\implies \ell _{x}(1-{}_{n}p_{x})=\ell _{x}-\ell _{x+n}\implies -\ell _{x}{}_{n}p_{x}=-\ell _{x+n}\implies \ell _{x}{}_{n}p_{x}=\ell _{x+n}.$$

$\Box$

There are two types of expectation of life: one is discrete and another is continuous, and they are called curtate-expectation-of-life and complete-expectation-of-life respectively.

Definition. (Complete-expectation-of-life) The complete-expectation-of-life of $(x)$, denoted by ${\overset {\circ }{e}}_{x}$, is $\mathbb {E} [T_{x}]$.

Definition. (Curtate-expectation-of-life) The curtate-expectation-of-life of $(x)$, denoted by $e_{x}$, is $\mathbb {E} [K_{x}]$.

Proposition. ${\overset {\circ }{e}}_{x}=\int _{0}^{\infty }{}_{t}p_{x}\,dt$.

Proof. We will use integration by parts. $${\overset {\circ }{e}}_{x}=\mathbb {E} [T_{x}]=\int _{0}^{\infty }tf_{x}(t)\,dt=\int _{0}^{\infty }t\,d(F_{x}(t))=\int _{0}^{\infty }t\,d(_{t}q_{x})=\int _{0}^{\infty }t\,d(1-{}_{t}p_{x})=\int _{0}^{\infty }t\,d(-{}_{t}p_{x})=[t(-{}_{t}p_{x})]_{0}^{\infty }-\int _{0}^{\infty }-{}_{t}p_{x}\,dt=\lim _{t\to \infty }(-t\;_{t}p_{x})+\int _{0}^{\infty }{}_{t}p_{x}\,dt$$ Now, it suffices to prove that $\lim _{t\to \infty }(-t\;_{t}p_{x})=0$ and this is true since $-t\to -\infty$ and $_{t}p_{x}=S_{x}(t)\to 0$ as $t\to \infty$, so this limit either equals $-\infty$ or 0. However, since the expected value exists (i.e. does not tend to infinity, or else the expectation of life does not make sense), this limit cannot equal $-\infty$, and so this limit is 0.

$\Box$

Remark.

• With a similar proof, we can also prove that $\mathbb {E} [T_{x}^{2}]{\overset {\text{ def }}{=}}\int _{0}^{\infty }t^{2}f_{x}(t)\,dt=\int _{0}^{\infty }{}_{t}p_{x}\,d({\color {darkgreen}t^{2}})={\color {darkgreen}2}\int _{0}^{\infty }{\color {darkgreen}t}\;_{t}p_{x}\,dt$.
• Thus, the variance $\operatorname {Var} (T_{x})=2\int _{0}^{\infty }t\;_{t}p_{x}\,dt-({\overset {\circ }{e}}_{x})^{2}$.

Proposition. $e_{x}=\sum _{k=1}^{\infty }{}_{k}p_{x}$.

Proof. The previous proposition about ${\overset {\circ }{e}}_{x}$ uses integration by parts in the proof, and we can analogously use summation by parts (may be interpreted as a discrete analogue of integration by parts) in the proof. However, there is a simpler way to prove this proposition, where the summation is "split" appropriately: $${\begin{aligned}e_{x}&=\mathbb {E} [K_{x}]\\&=\sum _{k=0}^{\infty }k{}_{k}p_{x}q_{x+k}\\&=\sum _{k=0}^{\infty }k({}_{k}p_{x}-{}_{k+1}p_{x})\\&=\sum _{k={\color {darkgreen}1}}^{\infty }k{}_{k}p_{x}-\sum _{k=0}^{\infty }k{}_{k+1}p_{x}&(k{}_{k}p_{x}=0{\text{ when }}k=0)\\&=\sum _{k={\color {darkgreen}1}}^{\infty }k{}_{k}p_{x}-\sum _{k'=1}^{\infty }(k'-1){}_{k'}p_{x}&(k'=k+1)\\&=\underbrace {\sum _{k={\color {darkgreen}1}}^{\infty }k{}_{k}p_{x}-\sum _{k'=1}^{\infty }k'{}_{k'}p_{x}} _{=0}+\sum _{k'=1}^{\infty }{}_{k'}p_{x}&({\text{the first two sums represent the same thing}})\\&=\sum _{k'=1}^{\infty }{}_{k'}p_{x}.\end{aligned}}$$ We can observe that this sum and the sum in the proposition represent the same thing, and thus the result follows.

Proof using summation by parts

The idea in this proof is similar to the proof for the proposition about ${\overset {\circ }{e}}_{x}$, except that the integration by parts is replaced by summation by parts (which can be interpreted as the "discrete version" of integration by parts). The proof for summation by parts is omitted here. The details is as follows: First, define the forward difference $\Delta f(x)=f(x+1)-f(x)$ for the summation by parts (this is analogous to "$dx$" in the integral). Then, $${\begin{aligned}e_{x}&=\mathbb {E} [K_{x}]\\&=\sum _{k=0}^{\infty }k({}_{k}p_{x})(q_{x+k})\\&=\sum _{k=0}^{\infty }k({}_{k}p_{x}-{}_{k+1}p_{x})\\&=\sum _{k=0}^{\infty }k\Delta (-{}_{k}p_{x})\\&=[k(-{}_{k}p_{x})]_{0}^{\infty }-\sum _{k=0}^{\infty }-{}_{\color {darkgreen}k+1}p_{x}\underbrace {\Delta (k)} _{=k+1-k=1}&{\text{by summation by parts formula}}\\&=\lim _{k\to \infty }(-k\;_{k}p_{x})+\sum _{\color {darkgreen}k=1}^{\infty }{}_{\color {darkgreen}k}p_{x}.\end{aligned}}$$ In particular, we have $_{k}p_{x}(q_{x+k})={\frac {S_{0}(x+k)}{S_{0}(x)}}\left(1-{\frac {S_{0}(x+k+1)}{S_{0}(x+k)}}\right)={\frac {S_{0}(x+k)}{S_{0}(x)}}-{\frac {S_{0}(x+k+1)}{S_{0}(x)}}={}_{k}p_{x}-{}_{k+1}p_{x}$.

Again, it suffices to prove that $\lim _{k\to \infty }(-k\;_{k}p_{x})=0$, and this is true for a similar reason: existence of the expected value $\mathbb {E} [K_{x}]$.

$\Box$

Remark.

• With a similar proof, we can also prove that $\mathbb {E} [K_{x}^{2}]=\sum _{k=0}^{\infty }k^{2}\Delta (-{}_{k}p_{x})=\underbrace {[k^{2}(-{}_{k}p_{x})]_{0}^{\infty }} _{=0}-\sum _{k=0}^{\infty }{}_{k+1}p_{x}\underbrace {\Delta ({\color {darkgreen}k^{2}})} _{=k^{2}+2k+1-k^{2}=2k+1}=\sum _{k=0}^{\infty }({\color {darkgreen}2k+1}){}_{k+1}p_{x}=\sum _{\color {darkgreen}k=1}^{\infty }({\color {darkgreen}2k-1}){}_{k}p_{x}$ (for summation by parts method).
• Thus, the variance $\operatorname {Var} (K_{x})=\left(\sum _{k=1}^{\infty }(2k-1){}_{k}p_{x}\right)-(e_{x})^{2}$.

Exercise. Prove that $\mathbb {E} [K_{x}^{2}]=\sum _{k=1}^{\infty }(2k-1){}_{k}p_{x}$ using the "sum splitting" method.

Solution

Proof. $${\begin{aligned}\mathbb {E} [K_{x}^{2}]&=\sum _{k=0}^{\infty }k^{2}{}_{k}p_{x}q_{x+k}\\&=\sum _{k=0}^{\infty }k^{2}({}_{k}p_{x}-{}_{k+1}p_{x})\\&=\sum _{k={\color {darkgreen}1}}^{\infty }k^{2}{}_{k}p_{x}-\sum _{k=0}^{\infty }k^{2}{}_{k+1}p_{x}&(k^{2}{}_{k}p_{x}{\text{ when }}k=0)\\&=\sum _{k={\color {darkgreen}1}}^{\infty }k^{2}{}_{k}p_{x}-\sum _{k'=1}^{\infty }(k'-1)^{2}{}_{k'}p_{x}&(k'=k+1)\\&=\underbrace {\sum _{k={\color {darkgreen}1}}^{\infty }k^{2}{}_{k}p_{x}-\sum _{k'=1}^{\infty }k'^{2}{}_{k'}p_{x}} _{=0}+\sum _{k'=1}^{\infty }(2k'-1){}_{k'}p_{x}&(-(k'-1)^{2}=-k'+2k'-1)\\&=\sum _{k'=1}^{\infty }(2k'-1){}_{k'}p_{x}.\end{aligned}}$$

$\Box$

The following are recursion relations for ${\overset {\circ }{e}}_{x}$ and $e_{x}$, which can be useful when we want to find the complete/curtate-expectation-of-life of $(x)$ given the expectation of a life with some other ages, say $x+1$ and $x+2$.

We will state the recursion relations as a form of proposition, and then prove them formally. After the proof, we will try to give some intuitive explanations about the recursion relation for $e_{x}$.

Proposition. $e_{x}=p_{x}(1+e_{x+1})$.

Proof. $$e_{x}=\sum _{k=1}^{\infty }{}_{k}p_{x}=p_{x}+\sum _{k=2}^{\infty }{}_{k}p_{x}=p_{x}+\sum _{k=2}^{\infty }p_{x}{}_{k-1}p_{x+1}=p_{x}+p_{x}\sum _{k=1}^{\infty }{}_{k}p_{x+1}=p_{x}(1+e_{x+1}).$$ In particular, we have $_{k}p_{x}={\frac {S_{0}(x+k)}{S_{0}(x)}}={\frac {S_{0}(x+1)}{S_{0}(x)}}\cdot {\frac {S_{0}(x+k)}{S_{0}(x+1)}}=p_{x}\;_{k-1}p_{x+1}$.

$\Box$

An intuitive explanation of this recursion relation is as follows:

• for LHS, $e_{x}$ is the curtate-expectation-of-life of $(x)$;
• for RHS, $e_{x+1}$ is the curtate-expectation-of-life of $(x+1)$, and we want to "transform" it to the expectation of $(x)$. The first step is adding 1 to it, since this is the expectation with respect to $(x+1)$, but we want the expectation from the perspective of $(x)$, which is 1 year younger. But only this step is not enough, since "$e_{x+1}$" assumes the life already lives for $x+1$ years, but for $e_{x}$, the life is only assumed to live for $x$ years. Hence, we also need to multiply the probability for $(x)$ to live for one year, $p_{x}$, to "get to" $e_{x+1}$.

• Now, "the expectation of life from age $x+1$ onward" is done through $e_{x+1}$. How about "the expectation of life from age $x$ to age $x+1$"? Indeed, when the life dies within age $x$ and $x+1$, $K=0$. This means such "expectation of life" is zero.

Proposition. ${\overset {\circ }{e}}_{x}=\int _{0}^{1}{}_{t}p_{x}\,dt+p_{x}{\overset {\circ }{e}}_{x+1}$.

Proof. $${\begin{aligned}{\overset {\circ }{e}}_{x}&=\int _{0}^{\infty }{}_{t}p_{x}\,dt\\&=\int _{0}^{1}{}_{t}p_{x}\,dt+\int _{1}^{\infty }{}_{t}p_{x}\,dt\\&=\int _{0}^{1}{}_{t}p_{x}\,dt+p_{x}\int _{1}^{\infty }{}_{t-1}p_{x+1}\,dt\\&=\int _{0}^{1}{}_{t}p_{x}\,dt+p_{x}\int _{\color {darkgreen}0}^{\infty }{}_{\color {darkgreen}u}p_{x+1}\,d{\color {darkgreen}u}&(u=t-1)\\&=\int _{0}^{1}{}_{t}p_{x}\,dt+p_{x}{\overset {\circ }{e}}_{x+1}.\\\end{aligned}}$$

$\Box$

Example. Given that $S_{0}(x)={\frac {50-x}{50}},\quad 0\leq x\leq 50$, we have ${\overset {\circ }{e}}_{x}=\int _{0}^{\infty }{}_{t}p_{x}\,dt=\int _{0}^{\infty }{\frac {S_{0}(x+t)}{S_{0}(x)}}\,dt=\int _{0}^{50-x}{\frac {(50-x-t)/50}{(50-x)/50}}\,dt=\int _{0}^{50-x}{\frac {50-x-t}{50-x}}\,dt=\int _{0}^{50-x}\left(1-{\frac {t}{50-x}}\right)\,dt=\left[t-{\frac {t^{2}}{100-2x}}\right]_{0}^{50-x}=\left(50-x-{\frac {(50-x)^{2}}{2(50-x)}}\right)={\frac {50-x}{2}}$. In particular, $S_{0}(x+t)$ has nonzero value when $0\leq x+t\leq 50$, and $S_{0}(x)$ has nonzero value when $0\leq x\leq 50$. Also, $_{t}p_{x}={\frac {S_{0}(x+t)}{S_{0}(x)}}$ holds only when $t\geq 0$. It follows that the bounds on $x$ and $t$ are given by $0\leq x\leq 50-t$ and $0\leq t\leq 50-x$.

Exercise. It is given that $S_{0}(x)=1-{\frac {x}{5}},\quad 0\leq x\leq 5$.

1 Calculate ${\overset {\circ }{e}}_{3}$.

	0.5
	0.75
	1
	1.25
	1.5

2 Calculate $e_{3}$.

	0.5
	0.75
	1
	1.25
	1.5

Assumptions for fractional ages

Previously, we have discussed the continuous random variable $T_{x}$ and discrete random variable $K_{x}$. A life table can specify the distribution of $K_{x}$ since the values of $q_{x}$ for different integer $x$ can be obtained from the life table. However, the life table is not enough to specify the distribution of $T_{x}$, since we do not know the value of $q_{x}$ when $x$ is not an integer. Thus, in order to specify a distribution of $T_{x}$ using a life table, we need to make some assumptions about the fractional (non-integer) ages.

In actuarial science, three assumptions are widely used, namely uniform distribution of deaths (UDD) (or linear interpolation), constant force of mortality (or exponential interpolation), and hyperbolic (or Balducci) assumption (or harmonic interpolation). We will define them each using survival functions, as follows:

Definition. (Uniform distribution of deaths) Uniform distribution of deaths assumption assumes $$S_{0}(x+t)=(1-t)S_{0}(x)+tS_{0}(x+1),\quad 0\leq t\leq 1.$$

Definition. (Constant force of mortality) Constant force of mortality assumption assumes $$\ln S_{0}(x+t)=(1-t)\ln S_{0}(x)+t\ln S_{0}(x+1),\quad 0\leq t\leq 1.$$

Remark.

• The equation can be alternatively stated as $S_{0}(x+t)=[S_{0}(x)]^{1-t}[S_{0}(x+1)]^{t},\quad 0\leq t\leq 1$.

Definition. (Balducci assumption) Balducci assumption assumes $${\frac {1}{S_{0}(x+t)}}={\frac {1-t}{S_{0}(x)}}+{\frac {t}{S_{0}(x+1)}},\quad 0\leq t\leq 1.$$

Remark.

• It is also called hyperbolic assumption.

Under UDD assumption, we have some "nice" and simple expressions for various probabilities related to mortality. We can obtain those expressions by substituting $S_{0}(x+t)$ by the RHS of the equation mentioned in the assumption.

Example. Under UDD assumption, we have, when $0\leq t\leq 1$, $$_{t}q_{x}=1-{\frac {S_{0}(x+t)}{S_{0}(x)}}=1-{\frac {(1-t)S_{0}(x)+tS_{0}(x+1)}{S_{0}(x)}}=1-(1-t)-t(p_{x})=t-t(p_{x})=t(1-p_{x})=t(q_{x}).$$

Exercise. Show that under UDD assumption, when $0\leq t\leq 1$, $\mu _{x+t}={\frac {q_{x}}{1-t(q_{x})}}$.

Answer

Under UDD assumption, when $0\leq t\leq 1$, $$\mu _{x+t}=-{\frac {{\frac {\partial }{\partial t}}[S_{0}(x+t)]}{S_{0}(x+t)}}=-{\frac {-S_{0}(x)+S_{0}(x+1)}{S_{0}(x+t)}}={\frac {S_{0}(x)-S_{0}(x+1)}{S_{0}(x+t)}}={\frac {{\big (}S_{0}(x)-S_{0}(x+1){\big )}/S_{0}(x)}{{\big (}S_{0}(x+t){\big )}/S_{0}(x)}}={\frac {q_{x}}{1-{}_{t}q_{x}}}={\frac {q_{x}}{1-t(q_{x})}}$$ In particular, we differentiate with respect to $t$ instead of $x$ since only $t$ is varying, and $x$ should be fixed at an selected integer. Also, we use the result in the previous example: $_{t}q_{x}=t(q_{x})$ in the last step.

For the three assumption mentioned, there is a particularly "nice" and simple result for each of them, and we may use those "nice" results for the calculation in practice, rather than applying the definitions. The "nice" result for UDD assumption is mentioned in a previous example: when $0\leq t\leq 1$, ${}_{t}q_{x}=t(q_{x})$. The "nice" results for the other two assumptions are as follows:

Theorem. (Notable property for constant force assumption) Under constant force assumption, $_{t}p_{x}=p_{x}^{t},\quad 0\leq t\leq 1$.

Proof. $$_{t}p_{x}={\frac {S_{0}(x+t)}{S_{0}(x)}}={\frac {[S_{0}(x)]^{1-t}[S_{0}(x+1)]^{t}}{S_{0}(x)}}={\frac {[S_{0}(x+1)]^{t}}{[S_{0}(x)]^{t}}}=\left({\frac {S_{0}(x+1)}{S_{0}(x)}}\right)^{t}=p_{x}^{t}.$$

$\Box$

Theorem. (Notable property for Balducci assumption) Under Balducci assumption, $_{1-t}q_{x+t}=(1-t)q_{x},\quad 0\leq t\leq 1$.

Proof. $$_{1-t}q_{x+t}=1-{\frac {S_{0}(x+1)}{S_{0}(x+t)}}=1-S_{0}(x+1)\left({\frac {1-t}{S_{0}(x)}}+{\frac {t}{S_{0}(x+1)}}\right)=1-(1-t){\frac {S_{0}(x+1)}{S_{0}(x)}}-t=1-t-(1-t)p_{x}=(1-t)(1-p_{x})=(1-t)q_{x}.$$

$\Box$

An interesting result under the UDD assumption is related to the independence of two random variables.

To simplify the notations, from now on, we let $T=T_{x}$ and $K=K_{x}$ unless otherwise specified.

Define a continuous random variable $S$ by $T=K+S,\quad S\in (0,1)$. That is, $S$ is the random variable representing the fractional part of a year lived in the year of death of $(x)$. For example, if $S=0.5$, then $(x)$ lives for half year in the year of death.

Then, $K$ and $S$ are independent under UDD assumption. This is because $$\mathbb {P} (K=k{\text{ and }}S\leq s)=\mathbb {P} (k\leq T\leq k+s)={}_{k}p_{x}{}_{s}q_{x+k}={}_{k}p_{x}\cdot s\cdot q_{x+k}=(_{k|}q_{x})\cdot s=\mathbb {P} (K=k)\mathbb {P} (S\leq s)$$ under UDD assumption. Also, we can observe that the cdf of $S$ is $\mathbb {P} (S\leq s)=s={\frac {s-0}{1-0}}$, which is the cdf of uniform distribution with support $(0,1)$. This means $S$ follows the uniform distribution on $(0,1)$ under UDD assumption. Hence, $\mathbb {E} [S]={\frac {1}{2}}$ and $\operatorname {Var} (S)={\frac {1}{12}}$. This gives rise to results under UDD assumptions:

• ${\overset {\circ }{e}}_{x}=\mathbb {E} [T]=\mathbb {E} [K+S]=\mathbb {E} [K]+\mathbb {E} [S]=e_{x}+{\frac {1}{2}}$.
• $\operatorname {Var} (T)=\operatorname {Var} (T){\overset {\text{ independence }}{=}}\operatorname {Var} (K)+\operatorname {Var} (S)=\left(\sum _{k=1}^{\infty }(2k-1){}_{k}p_{x}\right)-(e_{x})^{2}+{\frac {1}{12}}$.

These two results give us an alternative way to calculate the mean and variance of $T$ where only discrete things are used in the calculation. However, we should be careful that these results hold under UDD assumption, so we cannot use these results without UDD assumption.

Exercise. (Previous exercise revisit) From a previous exercise, we are asked to calculate ${\overset {\circ }{e}}_{3}$ and $e_{3}$ given $S_{0}(x)=1-{\frac {x}{5}},\quad 0\leq x\leq 5$. The following questions are some further questions on this exercise.

(a) Show that the UDD assumption is satisfied with this survival function.

(b) Verify that ${\overset {\circ }{e}}_{3}=e_{3}+0.5$.

Solution

(a)

Proof. Since when $t\in [0,1]$, $$S_{0}(x+t)=1-{\frac {x+t}{5}},$$ and $$(1-t)S_{0}(x)+tS_{0}(x+1)=(1-t)\left(1-{\frac {x}{5}}\right)+t\left(1-{\frac {x+1}{5}}\right)=1-{\frac {x}{5}}-t+{\frac {tx}{5}}+t-{\frac {tx}{5}}-{\frac {t}{5}}=1-{\frac {x+t}{5}},$$ the UDD assumption is satisfied with this survival function by definition.

$\Box$

(b) From the previous exercise, we have ${\overset {\circ }{e}}_{3}=1$ and $e_{3}=0.5$. Hence, ${\overset {\circ }{e}}_{3}=e_{3}+0.5$.

Laws of mortality

In this section, we will introduce some simple laws of mortality (i.e. some specified distributions for mortality). Some of these laws may be appropriate to model the human mortality for some ages, but it is commonly believed that none of these laws is appropriate to model the human mortality for all ages.

Indeed, if we want to model the human mortality using some probability distribution, we may need a mixture of distributions, since the mortality should be distributed in a different manner when human is in different ages, and thus different distributions should be used in different ages. To choose a suitable distribution for some ages, we may investigate the shape of corresponding empirical distribution based on the actual human mortality data, and pick a suitable distribution accordingly. For example, if the mortality increases exponentially for some ages, then we may select a distribution for which the force of mortality also increases exponentially.

In practice, to calculate the probabilities related to human mortality, we usually use life tables for the calculations. This is usually the case for insurance companies. Each insurance company has its own life table, based on the mortality data of, possibly its clients. Since such life table is constructed using the actual human mortality data based on past experiences, the life table is usually deemed to be more accurate than a specific distribution.

Nevertheless, having a law for mortality allows us to simplify the calculation of the probabilities related to mortality.

Definition. (de Moivre's law of mortality) The de Moivre's law of mortality has the force of mortality $\mu _{x}={\frac {1}{\omega -x}}$ and the survival distribution $S_{0}(x)=1-{\frac {x}{\omega }}$, with $0\leq x<\omega$ ($\omega$ is called the limiting age, i.e. an human must die (strictly) before that age).

Remark.

• This is the simplest law among the laws we discuss.
• Indeed, in this law, the mortality follows uniform distribution on $[0,\omega )$, since the cdf is ${\frac {x}{\omega }},\quad 0\leq x<\omega$, which is exactly the same as the cdf of uniform distribution on $[0,\omega )$.

• Hence, the pdf $f_{0}(x)={\frac {1}{\omega }},\quad 0\leq x<\omega$.

Exercise. Verify that if the force of mortality is $\mu _{x}={\frac {1}{\omega -x}}$ ($0\leq x<\omega$), then the survival distribution $S_{0}(x)=1-{\frac {x}{\omega }}$ ($0\leq x<\omega$).

Solution

When $\mu _{x}={\frac {1}{\omega -x}}$ where $0\leq x<\omega$, we have $$S_{0}(x)=\exp \left(-\int _{0}^{x}\mu _{t}\,dt\right)=\exp \left(-\int _{0}^{x}{\frac {1}{\omega -t}}\,dt\right)=\exp \left([\ln |\omega -t|]_{0}^{x}\right)=\exp(\underbrace {\ln(\omega -x)} _{\omega -x>0}-\ln \omega )=\exp \left(\ln \left({\frac {\omega -x}{\omega }}\right)\right)=1-{\frac {x}{\omega }}.$$

Example. Show that under de Moivre's law of mortality, ${}_{t}p_{x}=S_{x}(t)=1-{\frac {t}{\omega -x}}$ (where $0\leq x<\omega$), with certain bounds on $t$.

Proof. Under de Movire's law of mortality, we have $$S_{x}(t)={\frac {S_{0}(x+t)}{S_{0}(x)}}={\frac {1-{\frac {x+t}{\omega }}}{1-{\frac {x}{\omega }}}}={\frac {(\omega -x-t)/\omega }{(\omega -x)/\omega }}=1-{\frac {t}{\omega -x}}$$ with certain bounds on $t$.

$\Box$

Exercise.

1. Hence, show that $q_{x}=\mu _{x}$ under de Moivre's law of mortality.

Solution

Proof. By above, we have $p_{x}=1-{\frac {1}{\omega -x}}$ under de Moivre's law of mortality. It follows that $$q_{x}={\frac {1}{\omega -x}}=\mu _{x}.$$

$\Box$

2. What are the bounds on $x$ in the above example?

Solution

The bounds on $t$ are given by $0\leq t<\omega -x$, since for $S_{0}(x+t)$, the bounds are $0\leq x+t<\omega \implies -x\leq t<\omega -x$. But $t\geq 0$ since $t$ represents time. So, the bounds on $t$ are given by $0\leq t<\omega -x$.

3. Considering the above example, which distribution does $T_{x}$ follow?

Solution

From the above example, the cdf $F_{x}(t)={\frac {t}{\omega -x}},\quad 0\leq t<\omega -x$. So, $T_{x}$ follows the uniform distribution on $[0,\omega -x)$.

Remark.

• From this, we can see that when the age $x$ increases, the type of distribution does not change (still uniform), but the right endpoint of the support decreases.

4. Hence, show that ${\overset {\circ }{e}}_{x}={\frac {\omega -x}{2}}$.

Solution

Proof. Since $T_{x}$ follows the uniform distribution on $[0,\omega -x)$, it follows that the mean $${\overset {\circ }{e}}_{x}=\mathbb {E} [T_{x}]={\frac {\omega -x}{2}}.$$

$\Box$

Definition. (Gompertz's law of mortality) The Gompertz's law of mortality has the force of mortality $\mu _{x}=Bc^{x}$ and the survival distribution $S_{0}(x)=\exp \left(-{\frac {B}{\ln c}}(c^{x}-1)\right)$, with $B>0,c>1$ and $x\geq 0$.

Definition. (Makeham's law of mortality) The Makeham's law of mortality has the force mortality $\mu _{x}=A+Bc^{x}$ and the survival distribution $S_{0}(x)=\exp \left(-Ax-{\frac {B}{\ln C}}(c^{x}-1)\right)$, with $B>0,A\geq -B,c>1$ and $x\geq 0$.

Remark.

• The Makeham's law of mortality is a generalization of the Gompertz's law of mortality, since these two laws are exactly the same when $A=0$.

Definition. (Weibull's law of mortality) The Weibull's law of mortality has the force mortality $\mu _{x}=kx^{n}$ and the survival distribution $S_{0}(x)=\exp \left(-{\frac {kx^{n+1}}{n+1}}\right)$, with $k\geq 0,n>-1$ and $x\geq 0$.

Select and ultimate table

When a person purchases a life insurance policy offered by an insurance company, he needs to give some personal information to the insurance company, e.g. some information about his health status. For the insurance company to decide whether it should sell the policy to the person, those information provided by that person is accessed through the process of underwriting. For underwriting, the underwriter check the information to see whether the risk in insuring that person is appropriate.

Without underwriting, it is likely that people will only purchase life insurance policy when they think they will die soon (e.g. they have a very serious disease), so that they will likely have early claims. In this case, the insurance company may need to pay a large amount of money and suffer a great loss in a short time, and then bankrupt. This shows the necessity of underwriting.

Basically, the "select" in the section title arises from the underwriting process, and when we say an individual is "selected" at age $x$, he is underwritten at age $x$ (so the most recent information about the individual is known). Since there are some new information about the individual when he is underwritten (or selected), we will expect there is some update in his survival distribution, and therefore his probabilities related to mortality will change as well. Because of this, we need to have some changes in the actuarial notations, depending on the selection age.

In such actuarial notations, we usually add square brackets around the age at selection, and the numbers in the subscript change accordingly. For example, $q_{25}$ becomes $q_{[25]}$ if the selection age is 25, $q_{[12]+13}$ if the selection age is 12.

Since when a person is underwritten long time ago, he may have poorer health condition (e.g. getting older having some new diseases) from the time at which he is underwritten to now, we will intuitively expect that the longer time passed from the time at which a person aged $x$ is underwritten, the more likely that person will be die in the coming year. That is, $q_{[0]+x}>q_{[1]+x}>\dotsb >q_{[x-1]+1}>q_{[x]}$ .

The impact on the survival distribution from the selection age may decrease when the time passed from the selection is longer. Beyond a certain time period, say $r$ years, the "$q$"'s at the same attained age (i.e. selection age plus the time passed from it to now) but different selection ages will be very close. In other words, $q_{[x-j]+r+j}\approx q_{[x]+r},\quad 0\leq j\leq x$ (the condition on $j$ is to ensure that the selection age at LHS $x-j\geq 0$) . Such $r$ years is called the select period. Because the "$q$"'s mentioned above will be very close, all such "$q$"'s will only be written as $q_{x+r}$, without any square brackets (since the effect of selection is basically "gone", and so the square brackets are gone). For example, if the selection period is 2 years, then $q_{[10]+2}$ and $q_{[5]+7}$ will be both written as $q_{12}$ simply. However, we will not write $q_{[15]+1}$ and $q_{[16]}$ as $q_{16}$, since these two "$q$"'s are "quite different" because of the selection impact is still "quite large".

The following are some terms related to life table.

• An aggregate table is a life table in which the functions are only given for attained ages.
• A select table is a life table in which some function involves the age at selection.
• An ultimate table is usually appended to a select table as a last column to reflect the setting of select table. The combination of a select table and an ultimate table in such a way is called a select-and-ultimate table.

For example, an excerpt of select-and-ultimate table with select period 2 years may look like:

Select-and-ultimate table with 2-year select period

Age at selection, $x$	$q_{[x]}$	$q_{[x]+1}$	${\cancel {q_{[x]+2}}}q_{x+2}$
0	$q_{[0]}$	$q_{[0]+1}$	$q_{2}$
1	$q_{[1]}$	$q_{[1]+1}$	$q_{3}$
2	$q_{[2]}$	$q_{[2]+1}$	$q_{4}$
...	...	...	...

The last column is the ultimate table. We can observe that we do not need additional columns for $q_{x+3},q_{x+4}$ etc., since we can already get such values in the "$q_{x+2}$" value in a different row (with different $x$).

Given a select-and-ultimate table, we can do various calculations based on it.

Example. You are given the following except of a (hypothetical) select-and-ultimate table with 4-year select period: $${\begin{array}{ccccccc}x&\ell _{[x]}&\ell _{[x]+1}&\ell _{[x]+2}&\ell _{[x]+3}&\ell _{x+4}\\\hline 13&9910&9905&9890&9855&9800\\14&9820&9800&9790&9750&9730\\15&9750&9700&9685&9650&9620\\\end{array}}$$ Then, $p_{[13]}={\frac {\ell _{[13]+1}}{\ell _{[13]}}}\approx 0.999495$, $_{2}q_{[14]}=1-{\frac {\ell _{[14]+2}}{\ell _{[14]}}}=1-{\frac {9790}{9820}}\approx 0.010183$, and $_{6}p_{[13]}={\frac {\ell _{19}}{\ell _{[13]}}}={\frac {9620}{9910}}\approx 0.9707$ (since the select period is 4 years).

Exercise.

Which of the following probability (or probabilities) cannot be calculated from the above select-and-ultimate table?

	$_{4}q_{[12]}$
	$_{5}p_{[14]}$
	$p_{[15]}$
	$_{2|3}q_{[15]}$